# How do you solve 53^(x + 1) = 65.4?

Apr 22, 2016

$x \approx 0.053$

#### Explanation:

First the the $\log$ of both sides:
${53}^{x + 1} = 65.4$
$\log {53}^{x + 1} = \log 65.4$
Then because of the rule $\log {a}^{b} = b \log a$, we can simplify and solve:
$\left(x + 1\right) \log 53 = \log 65.4$
$x \log 53 + \log 53 = \log 65.4$
$x \log 53 = \log 65.4 - \log 53$
$x = \frac{\log 65.4 - \log 53}{\log} 53$
And if you type this into your calculator you get:
$x \approx 0.053$